Question: Solve for $x$ : $ 2|x + 3| - 10 = -6|x + 3| + 9 $
Add $ {6|x + 3|} $ to both sides: $ \begin{eqnarray} 2|x + 3| - 10 &=& -6|x + 3| + 9 \\ \\ { + 6|x + 3|} && { + 6|x + 3|} \\ \\ 8|x + 3| - 10 &=& 9 \end{eqnarray} $ Add ${10}$ to both sides: $ \begin{eqnarray} 8|x + 3| - 10 &=& 9 \\ \\ { + 10} &=& { + 10} \\ \\ 8|x + 3| &=& 19 \end{eqnarray} $ Divide both sides by ${8}$ $ \dfrac{8|x + 3|} {{8}} = \dfrac{19} {{8}} $ Simplify: $ |x + 3| = \dfrac{19}{8}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 3 = -\dfrac{19}{8} $ or $ x + 3 = \dfrac{19}{8} $ Solve for the solution where $x + 3$ is negative: $ x + 3 = -\dfrac{19}{8} $ Subtract ${3}$ from both sides: $ \begin{eqnarray} x + 3 &=& -\dfrac{19}{8} \\ \\ {- 3} && {- 3} \\ \\ x &=& -\dfrac{19}{8} - 3 \end{eqnarray} $ Change the ${ - 3}$ to an equivalent fraction with a denominator of $8$ $ x = - \dfrac{19}{8} {- \dfrac{24}{8}} $ $ x = -\dfrac{43}{8} $ Then calculate the solution where $x + 3$ is positive: $ x + 3 = \dfrac{19}{8} $ Subtract ${3}$ from both sides: $ \begin{eqnarray} x + 3 &=& \dfrac{19}{8} \\ \\ {- 3} && {- 3} \\ \\ x &=& \dfrac{19}{8} - 3 \end{eqnarray} $ Change the ${ - 3}$ to an equivalent fraction with a denominator of $8$ $ x = \dfrac{19}{8} {- \dfrac{24}{8}} $ $ x = -\dfrac{5}{8} $ Thus, the correct answer is $x = -\dfrac{43}{8} $ or $x = -\dfrac{5}{8} $.